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Differential Equation

  1. Mar 31, 2008 #1
    dy/dx = (3x^2)(e^-y)

    1/(e^-y) dy = 3x^2 dx
    1/(e^-y) = x^3 + C

    Can anybody please give me any help or advice on to where to go from here? I'm having a really tough time with this one.
     
  2. jcsd
  3. Mar 31, 2008 #2
    [tex]\frac{dy}{dx}=3x^{2}e^{-y}[/tex]

    this diff. eq. can be solved with separation of variables.So:

    [tex]e^{y}dy=3x^{2}dx[/tex] now integrate both sides and solve for y.
     
  4. Mar 31, 2008 #3
    So, y = ln(x^3+C)?
     
  5. Mar 31, 2008 #4
    Looks like it is right.
     
  6. Mar 31, 2008 #5
    Thanks stupidmath!!! Now, if I wanted to solve, for example y(0)=1

    ln (C) = y - ln(x^3)
    ln (C) = 1 - 0
    C = e^1 ?

    y = ln(x^3) + e^1 ?
     
  7. Mar 31, 2008 #6
    Well, to impose the initial condition y(0)=1, you actually here have that when x=0, y=1.

    So it means:

    [tex]1=ln(c)=>C=e[/tex]

    so our particular solution would be

    [tex]y=ln(x^{3}+e)[/tex]

    You just misplaced the parentheses on your final answer.
     
  8. Mar 31, 2008 #7
    Ah, thanks a bunch stupidmath!!!
     
  9. Mar 31, 2008 #8
    One last question stupidmath. wouldn't dy/dx = 4(2y-1) end up being y = Ce^-32x?
     
  10. Apr 1, 2008 #9

    HallsofIvy

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    It is not "stupidmath". Beware of his ire!

    You should be able to do this one yourself now. If dy/dx= 4(2y-1) then dy/(2y-1)= 4dx.
    Integrating on both sides (1/2)ln|2y-1|= 4x+ C so ln|2y-1|= 8x+ C' (C'= 2C). Then 2y-1= C"e^(8c).
     
  11. Apr 1, 2008 #10

    arildno

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    It is suttypud-math, not stupidmath!
     
  12. Apr 1, 2008 #11
    Oh, I apologize for the faux pas sutupidmath. I also appreciate the help, thanks.
     
  13. Apr 1, 2008 #12
    Is there any meaning to this?:smile:
     
  14. Apr 2, 2008 #13

    Mute

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    Remember that ln(a + b) =/= ln(a) + ln(b), which is what you did here.
     
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